使用uci数据集中的data_banknote_authentication数据进行逻辑回归。 数据集：
代价cost:
最终目标是要求出使得 J(theta)最小时theta的值。采取的方法均为类似梯度下降法的方法。 不调库代码实现：
import pandas as pd # 加载数据集 data= pd.read_csv('D:/Desktop/data_banknote_authentication.txt',header=None) print(data.head()) X = data[[0,1,2,3]] # 取前四个作为特征 y = data[[4]] import math MAX_FEATURE_DIMENSION = 1024 MAX_SAMPLE_NUMBER = 1024 MAX_ITERATE_NUMBER = 1024 ##估价函数 def sigmoid(z): return(1 / (1.0 + math.exp(-z))) def hypothesis(x, theta, feature_number): h = 0.0 for i in range(feature_number+1): h += x[i] * theta[i] return(sigmoid(h)) ##计算偏导数 def compute_gradient(x, y, theta, feature_number, feature_pos, sample_number): sum = 0.0 for i in range(sample_number): h = hypothesis(x[i], theta, feature_number) sum += (h - y[i]) * x[i][feature_pos] return(sum / sample_number) ##代价 def compute_cost(x, y, theta, feature_number, sample_number): sum = 0.0 for i in range(sample_number): h = hypothesis(x[i], theta, feature_number) sum += -y[i] * math.log(h) - (1 - y[i]) * math.log(1 - h) return(sum / sample_number) ##梯度下降 def gradient_descent(x, y, theta, feature_number, sample_number, alpha, count): for i in range(count): tmp = [] for j in range(MAX_FEATURE_DIMENSION): tmp.append(0) for j in range(feature_number + 1): tmp[j] = theta[j] - alpha * compute_gradient(x, y ,theta, feature_number, j, sample_number) for j in range(feature_number + 1): theta[j] = tmp[j] print(compute_cost(x, y, theta, feature_number, sample_number)) if __name__ == '__main__': sample_number = X.shape[0] count = 30000 # 迭代次数 feature_number = 3 alpha = 0.002 # 步长 theta = [] for i in range(MAX_FEATURE_DIMENSION): theta.append(0) gradient_descent(X.values.tolist(), y[4].tolist(), theta, feature_number, sample_number, alpha, count) theta1 = [] for i in range(feature_number+1): theta1.append(theta[i]) print("cost收敛时的theta值：",theta1)代码实现（调用sklearn）：
import pandas as pd # 加载数据集 data= pd.read_csv('D:/Desktop/data_banknote_authentication.txt',header=None) print(data.head()) from sklearn.model_selection import train_test_split X = data[[0,1,2,3]] # 取前四个作为特征 y = data[[4]] X_train, X_test, y_train, y_test = train_test_split(X,y) X_train.head() from sklearn.linear_model import LogisticRegression clf = LogisticRegression(random_state=0, solver='lbfgs') ##在训练集上训练逻辑回归模型 clf.fit(X_train, y_train) ##查看其对应的w print('the weight of Logistic Regression:', clf.coef_) ##查看其对应的w0 print('the intercept(w0) of Logistic Regression:', clf.intercept_) ##在训练集和测试集上分别利用训练好的模型进行预测 train_predict = clf.predict(X_train) test_predict = clf.predict(X_test) from sklearn import metrics # import matplotlib.pyplot as plt ##利用accuracy（准确度）【预测正确的样本数目占总预测样本数目的比例】评估模型效果 print('The accuracy of the Logistic Regression is:', metrics.accuracy_score(y_train, train_predict)) print('The accuracy of the Logistic Regression is:', metrics.accuracy_score(y_test, test_predict)) ##查看混淆矩阵(预测值和真实值的各类情况统计矩阵) confusion_matrix_result = metrics.confusion_matrix(test_predict, y_test) print('The confusion matrix result:\n', confusion_matrix_result) # ##利用热力图对于结果进行可视化 # import seaborn as sns # plt.figure(figsize=(8, 6)) # sns.heatmap(confusion_matrix_result, annot=True, cmap='Blues') # plt.xlabel('Predictedlabels') # plt.ylabel('Truelabels') # plt.show()
使用uci数据集地址：http://archive.ics.uci.edu/ml/datasets/Combined+Cycle+Power+Plant
代价cost： 不调库代码实现（结果可能有点问题，需要对数据进行归一化）：
#每小时平均环境变量温度（T），环境压力（AP），相对湿度（RH）和排气真空（V），以预测工厂的每小时净电能输出（PE）。9568个数据 import pandas as pd # 加载数据集 data= pd.read_csv('D:/Desktop/Folds5x2_pp.CSV') data.columns=[0,1,2,3,4] print(data.head()) X = data[[0,1,2,3]] y = data[[4]] X.shape import math MAX_FEATURE_DIMENSION = 10000 MAX_SAMPLE_NUMBER = 10000 MAX_ITERATE_NUMBER = 10000 ##求导 def compute_gradient(x,y,theta,feature_number,feature_pos,sample_number): sum = 0.0 for i in range(sample_number): res = 0.0 for j in range(feature_number+1): res += x[i][j] * theta[j] sum += (res - y[i])*x[i][feature_pos] return sum/sample_number ##估价函数 def compute_cost(x,y,theta,feature_number,sample_number): sum = 0.0 for i in range(sample_number): res = 0.0 for j in range(feature_number+1): res += x[i][j] * theta[j] sum += (res - y[i]) * (res - y[i]) return sum/(2*sample_number) ##梯度下降法 def gradient_descent(x,y,theta,feature_number,sample_number,alpha,count): for i in range(count): tmp = [] for j in range(MAX_FEATURE_DIMENSION): tmp.append(0) for j in range(feature_number+1): tmp[j] = theta[j] - alpha * compute_gradient(x,y,theta,feature_number,j,sample_number) for j in range(feature_number+1): theta[j] = tmp[j] print("第",i,"次",compute_cost(x,y,theta,feature_number,sample_number)) ##测试 def predict(theta,x,feature_number): sum = 0.0 for i in range(feature_number+1): sum += theta[i]*x[i] return sum if __name__ == '__main__': sample_number = X.shape[0] alpha = 0.0001 count = 1500 feature_number = 3 theta = [] for i in range(MAX_FEATURE_DIMENSION): theta.append(0) gradient_descent(X.values.tolist(),y[4].tolist(),theta,feature_number,sample_number,alpha,count) print("[28.65,55.4,1000.25,40,95]的预测值：",predict(theta, [28.65,55.4,1000.25,40,95], feature_number))代码实现（调用sklearen）：
#每小时平均环境变量温度（T），环境压力（AP），相对湿度（RH）和排气真空（V），以预测工厂的每小时净电能输出（PE）。9568个数据 import pandas as pd # 加载数据集 data= pd.read_csv('D:/Desktop/Folds5x2_pp.CSV') data.columns=[0,1,2,3,4] X = data[[0,1,2,3]] y = data[[4]] from sklearn.model_selection import train_test_split X_train, X_test, y_train, y_test = train_test_split(X, y ,test_size = 1/3.,random_state = 8) from sklearn.linear_model import LinearRegression from sklearn.model_selection import cross_val_score # 线性回归 lr = LinearRegression(normalize=True,n_jobs=2) scores = cross_val_score(lr,X_train,y_train,cv=10,scoring='neg_mean_squared_error') #计算均方误差 print(scores.mean()) lr.fit(X_train,y_train) lr.score(X_test,y_test) # 多项式回归 from sklearn.preprocessing import PolynomialFeatures for k in range(1,4): lr_featurizer = PolynomialFeatures(degree=k) # 用于产生多项式 degree:最高次项的次数 print('-----',k,'-----') X_pf_train = lr_featurizer.fit_transform(X_train) X_pf_test = lr_featurizer.transform(X_test) pf_scores = cross_val_score(lr,X_pf_train,y_train,cv=10,scoring='neg_mean_squared_error') print(pf_scores.mean()) lr.fit(X_pf_train,y_train) print(lr.score(X_pf_test,y_test)) print(lr.score(X_pf_train,y_train)) #当 k=1 时，回归为线性回归； #当 k=2 时，回归为多项式回归，模型的效果比线性回归要好一点。 #当 k=3 时，已经出现过拟合现象，模型在训练集上得分很高，然而在验证集上得分很低，从交叉验证的得分结果也可以看得出来 # 正则化解决k=3的过拟合现象 lr_featurizer = PolynomialFeatures(degree=3) # 用于产生多项式 degree:最高次项的次数 X_pf_train = lr_featurizer.fit_transform(X_train) X_pf_test = lr_featurizer.transform(X_test) from sklearn.linear_model import Ridge # 岭回归 for a in [0,0.005]: print('----(%f)-----'%(a)) ridge = Ridge(alpha=a,normalize=True) pf_scores = cross_val_score(ridge,X_pf_train,y_train,cv=10,scoring='neg_mean_squared_error') print(pf_scores.mean()) ridge.fit(X_pf_train,y_train) print(ridge.score(X_pf_test,y_test)) print(ridge.score(X_pf_train,y_train)) # 正则化解决k=3的过拟合现象 lr_featurizer = PolynomialFeatures(degree=3) # 用于产生多项式 degree:最高次项的次数 X_pf_train = lr_featurizer.fit_transform(X_train) X_pf_test = lr_featurizer.transform(X_test) # LASSO回归： from sklearn.linear_model import Lasso for a in [0,0.0005]: print('----(%f)-----'%(a)) lasso = Lasso(alpha=a,normalize=True) pf_scores = cross_val_score(lasso,X_pf_train,y_train,cv=10,scoring='neg_mean_squared_error') print(pf_scores.mean()) lasso.fit(X_pf_train,y_train) print(lasso.score(X_pf_test,y_test)) print(lasso.score(X_pf_train,y_train))线性回归运行截图： 多项式回归： 正则化解决k=3的过拟合现象：
岭回归： LOSS回归：
